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# Mathematics Keynotes 2

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### Transcript of "Mathematics Keynotes 2"

1. 1. 1449/1 – 1 hour 15 minutes 40 objective questions 1449/2 – subjective Section A - 11 compulsory questions Section B - 4 out of 5 questions 2 Hours 30 minutes SCIENTIFIC CALCULATOR GEOMETRIC AL SET
2. 2. 3 marks
3. 3. Example ξ B A C II III IV V VI I Shade the set ( A ∩ B ) ∪ C 1. Label each part with a Roman number
4. 4. ξ B A C II III IV V VI I 2. Identify the shaded region U (A B) U C U ( II, III, IV III, IV, V, VI ) U IV, V
5. 5. 3. Identify the shaded region U ( II, III, IV III, IV, V, VI ) U IV, V III, IV U IV, V III, IV, V 4. Shade the region mark with III, IV, V U (A B) U C
6. 6. SETS Shade the Region (a) P ∩ R ' (b) P ∩ Q ∪ R' Q Q P P R Intersection R Union Com plim of ent
7. 7. P ∩ R' P ∩ Q ∪ R'
8. 8. Linear inequalities Shade the region that satisfies the inequalities Know how to sketch a straight line Application of y-intercept Understand the inequality sign < , > for dashed line and ≤ , ≥ for solid line
9. 9. Shade the region that satisfies the three inequalities y ≤ 2 x + 8 , y ≥ x and y < 8 y=x 0 y = 2x + 8
10. 10. Know how to sketch the straight line y =8 From , y = 2 x + 8 , y-intercept = 8 y = 2x + 8 y=x 8 y<8 Full marks
11. 11. y = 2x + 8 y=x 8 less marks
12. 12. 4 marks
13. 13. Solid Geometry & Volume Combination of two solids 1. Determine the two solids involved 2.Choose the operations + or - 3.Write the correct formulae 4.Substitute the values of r, h, d 22 Use π= 7
14. 14. Try this A hemisphere PQR has taken out from the cylinder. Find the volume of the remaining solid. P R Cylinder - hemisphere 2 3 πr h − πr 2 3 Q 8 cm 22 2 22 ×5× 5×8 − × 5× 5× 5 7 3 7 2 366 10 cm 3
15. 15. The diagram shows a solid cone with radius 9 cm and height 14 cm. A cylinder with radius 3 cm and height 7 cm is taken out of the solid. Calculate the volume, in cm3 , of the remaining solid. 22 Use π = 7
16. 16. Diagram 3 shows a solid cone with radius 9 cm and height 14 cm. A cylinder with radius 3 cm and height 7 cm is taken out of the solid. 14 cm Write the formula first 7 cm 1 2 Vcone = πr 3 cm 3 9 cm Vcylinder = π r h 2
17. 17. 1 Vcone = πr 2 h 3 Stress on correct values when substituting 1  22  ( 9 ) ( 14 ) 2 =  3 7  Remaining solid =1188 = 1188 – 198 Vcylinder =πr h 2 = 990  22  Stress on correct values ( 3 ) ( 7 ) 2 = when substituting  7  =198
18. 18. Lines and Planes in 3D 4 marks
19. 19. Lines and Planes in 3D IMPORTANT NOTES : 3. SKETC therig a le H ht- ng d tria le ng 2 Id ntify thea lea na eit . e ng nd m n g le b e t w e e n a lin e a n d a p la A n g le b e t w e e n t w o p la n e s
20. 20. LINES AND PLANES IN 3 DIMENSION Name the angle between the line CE and the plane EFGH Arrange the line and plane in two rows Find out the same alphabet Look at the line and the plane in the diagram C Write C in the first box B Look at c, choose Which One is the Nearest to C A D (slashed alphabet) 6 C E Look at C, WON E F G Choose W O N H (Non-slashed alphabets) F G Look at the diagram θ 6 C E C E G 6 H Tan θ= 6 Draw 3 boxes K2 √72 θ E 36 + 36 G θ= 35.26°
21. 21. Name the angle between the plane DGK and the base DEFG WON DK = GKW O N Slash the same alphabet Write K in the first box So, choose the midpoint of DG H Look at K, choose W O N Slashed- alphabet K D G K 8 G D E F G N θ 6 Look at the diagram D F Look at the diagram Look at the diagram 12 M K N M E Look at K, choose W O N K Tan θ= 6 EK = FK Non-slashed alphabets choose M 12 θ= 26.57° θ N M
22. 22. V WON WON Non-slash alphabet Slash alphabet 6 D C θ 8 A 10 B Calculate the angle between the plane AVC and the plane BVC A V C A Tan θ = 10 B V C 8 A C B θ= 51.34° θ C B
23. 23. A B Name the angle between the WON E F plane ACGE with the plane WON DCGH A C G E D θ C D C G H H G A C D between A & E, choose either one and write the alphabet in the first box ( for rectangle only)
24. 24. Try this.. H U G E F T P N L M R Name the angle between the plane LUM with the plane LPNM Answer : URT
25. 25. Find the angle between the plane JFE and the plane DEF. L J 5 F D 5 M 13 5 E
26. 26. L J 5 F D 5 M 13 5 E Identify the angle ∠ JMD
27. 27. 5 tan JMD = 13 − 5 2 2 ∠ JMD = 22.62 o o ' atau 22 37
28. 28. 5M A R K S
29. 29. MATHEMATICAL REASONING Is the following sentence a statement ? Give your reason. 9 + 2 = 2 −9 Statement yes Not accepted answer A statement. It is a false statement.
30. 30. Make a conclusion for the number sequence below 2 −1= 0 0 F u ll m a r k s 2 −1=1 1 3 d o ts 2 −1= 3 2 2 −1= 7 3 2 − 1, n = 0 ,1, 2 , 3 , ... n
31. 31. n 2 ─ 1 , n = 0 , 1 , 2 , 3 , .. n 2 ─ 1 , n =0,1,2,3, n 2 ─ 1 , n =0,1,2, n 2 ─ 1 Ie s s marks
32. 32. Write two implications from this compound statement 3 p = 5 if and only if p = 125 If 3 p = 5 , then p = 125 F u ll m a r k If p = 125 , then 3 p =5 3 p = 5 , then p = 125 n o m a rk If p = 125 , 3 p =5
33. 33. Complete the following argument Premise1 : If 4x = 16 , then x = 4 Premise 2 : x ≠ 4 Conclusion : 4 x ≠ 16
34. 34. SIMULTANEOUS LINEAR EQUATIONS Elimination method Substitution method Matrix method 4 M AR KS
35. 35. Solve the simultaneous linear equations 1 p − 2q = 9 3 5 p + 6q = −9 When there is a fraction, you must have the same denominator first p − 6q =9 3 p − 6q = 27
36. 36. Elimination method 1 p − 6q p − 2q = 9 same deno min ator =9 3 3 ⇒ p −6q = 27 1 5 p + 6q = −9 2 1 +2 6 p = 18 r r e c t o p e r a t io n ∴ p=3 ⇒ 3 − 6q = 27 ∴ q = −4
37. 37. Substitution method 1 p − 6q p − 2q = 9 same deno min ator =9 3 3 ⇒ p −6q = 27 ∴ p = 27 +6q C o r r e c t o p e r a t io n 1 5 p + 6q = −9 2 substitute 1 into 2 5 ( 27 + 6q ) + 6q = − 9 ⇒ p = 27 + 6 (− 4 ) 135 + 30q + 6q = −9 ∴ p=3 ∴ q = −4
38. 38. Matrix method 1  p  9   − 2  =   3  q   − 9  5 6     C o r r e c t m a t r ix f o r m  p 1  6 2 9   =    1   q  1 − 5 − 9    × 6 − 5 × ( − 2)  3 3  p  3   =   ∴ p = 3 , q = −4  q   −4    
39. 39. THE STRAIGHT LINE – 6 MARKS REMEMBER : y1 − y2 • Gradient m = x1 − x2 • Equation of a line y = mx + c x y + = 1 a b
40. 40. THE STRAIGHT LINE REMEMB ER : 3. Parallel lines , same gradient m1 = m2 4. Perpendicular lines , the product of their gradients = − 1 m1m2 = −1
41. 41. THE STRAIGHT LINE REMEMB ER : 5. x-intercept , substitute y=0 6. y-intercept , substitute x=0
42. 42. Important notes x-intercept = − 12 x = − 12 x-intercept is ( − 12 , 0 )
43. 43. In Diagram 2, O is the origin, point R lies on the x- axis and point P lies on the y-axis. Straight line PU is parallel to the x-axis and straight line PR is parallel to the straight line ST. The equation of straight line PR is x + 2y = 14. y (a) Find the value of its • • U P y-intercept from x + 2y = 14. x + 2y = 14 S • (b) Find the equation of the O • R x straight line ST and hence, state its x-intercept. • T (2,-5)
44. 44. (a) PU is parallel to the x-axis. Find the value of its y-intercept from x + 2y = 14. y P • • U 2 y = − x + 14 x 14 x + 2y = 14 y=− + S 2 2 • • x O R x y = − +7 2 • T (2,-5) y-intercept! y-intercept =7
45. 45. Find the equation of the straight line ST and hence, state its x-intercept. y From part (a), we have • • U x P 2 y = − x + 14 → y = − + 7 2 1 x + 2y = 14 Therefore, gradient ST , m = − S 2 • 1 • x Substitute m = − and po int (2, −5) in O R 2 y = mx + c,  1 −5 =  −  (2) + c • T (2,-5)  2 c = −5 + 1 = −4 Thus, equation ST is 1 y = − x−4 2
46. 46. hence, state its x-intercept. 1 y = − x−4 2 x − int ercept ∴ y = 0 1 4=− x 2 x = −8 x-intercept = – 8
47. 47. QUADRATIC EQUATIONS 4 MARKS
48. 48. QUADRATIC EQUATION REARRANGE TO GENERAL FORM OF QUADRATIC EQUATION ax + bx + c = 0 2 Factorise ( )( )=0 State the values of x
49. 49. Solve the quadratic equation 3n + 6n = 7 ( 1 − 2n ) 2 3n + 6n = 7 (1 − 2n ) ⇒ 3n + 6n − 7 + 14n = 0 2 2 ∴ 3n + 20n − 7 = 0 2 Fa c to rs mus t be g iv e n b y u s in g w h o le ( 3n − 1 ) ( n + 7 ) = 0 n u m b e rs 1 ∴ n= , −7 3
50. 50. 2k − 5 2 Solve the quadratic equation = 3k 3 2k − 5 = 9k 2 2k − 9k − 5 = 0 2 ( 2k + 1) ( k − 5) = 0 1 k =− , k =5 2
51. 51. MATRICES NOTES 1. When the matrix has no inverse ad − bc = 0 2. MATRIX FORM 3. Formula of the inverse matrix 4. State the value of x and of y
52. 52.  5 − 9 The inverse of matrix   2 − 3  is    1  − 3 9   q 5  p  Find the values of p and q
53. 53. Use the inverse formula 1  − 3 9 inverse =    − 2 5 5 × (−3) − (−9) × 2   1  − 3 9 Compare with the =   − 2 5  given inverse matrix 3  1  − 3 9 ∴ p= 3   q 5  p  q= −2
54. 54. Calculate the value of x and the value of y by using matrix method 2 x − 3 y = −14 − x + 3 y = 13 Form a matrix equation  2 − 3  x   − 14    − 1 3   y  =  13          
55. 55. Write the inverse formula IN FRONT 1  d − b   − c a  F u ll m a r k s ad − bc    x  1  3 3   − 14   =    y  3  1 2   13         x   − 1 ∴ x = −1  =    y  4      y=4
56. 56.  2 − 3  x   − 14     =   −1 3   y   13         x  1  3 3   − 14   =   y  3  1 2   13          x   − 1  =    y  4      le s s m a r k s
57. 57. Wrong arrangement  x   2 − 3  − 14     y   − 1 3  =  13           x  1  2 3  − 14   =    y  3  1 3  13         x  1  3 − 3   − 14   =   y  3  − 1 2   13    n o m a rk      x  − 14  1  3 3   =   y  13  3  1 2          
58. 58. CIRCLES : Perimeter and Area 1. Use the correct formulae 3. Substitute with the correct values. 22 π = 7
59. 59. P S 9 cm 7 cm W U 6 cm T 30O R Q 6 cm V 6 cm a) Find the perimeter of the shaded region b) Calculate the area of the shaded region
60. 60. P S 9 cm 7 cm W U 6 cm T 30O R Q 6 cm V 6 cm Perimeter = PU + UV + VQ + ( PQ – ST ) + SWT Area = area of triangle – area of hemisphere – area of the sector
61. 61. (a). Perimeter = PU + UV + VQ + ( PQ – ST ) + SWT  30 22   180 22  7   = 9+ × 2 × × 6  + 6 + ( 9 − 7) +   360 × 2 × 7 ×  2     360 7     218 = 7 1 31 7 31.14
62. 62. (b). Area 1    180 22  7  2  30 22 2   =  × 12 × 9  − × ×  − × × 6  2   360 7  2   360 7   709 = 28 9 25 28 25.32
63. 63. 1    180 22  7  2  30 22 2   a) =  × 12 × 9  −  × ×  − × × 6  2   360 7  2   360 7  709 = A n s w e r s a r e in t h e w ro n g 28 a ns w e r s pa c e s b)  30 22   180 22  7   = 9+ × 2 × × 6  + 6 + ( 9 − 7) +   360 × 2 × 7 ×  2     360 7     218 = 7
64. 64. Diagram 4 shows two sectors ORST and OUV with the same centre O. RWO is a semicircle with diameter RO and RO=2OV. ROV and OUT are straight lines. OV=7cm and angle UOV= 60˚ Calculate (a) perimeter of the whole diagram , (b) area of the shaded region. S T 7 c U W m o 6 0V R O 14 c 7 c m m
65. 65. a ) perimeter = RO + OV + RST + TU + UV  120 22  =14 + 7 +  × 2 × ×14  + 7  360 7 Stress on the correct values when substituting   60 22  + × 2× × 7  360 7  2 = 64 3
66. 66. b) area = OUV + ORST − OWR  60 22 2   120 22 2 = × ×7  +  × × 14   360 7   360 7  Stress on the correct values when substituting  180 22 2  − × ×7   360 7  = 154
67. 67. 4 marks
68. 68. PROBABILITY II n( A ) P( A) = n( S )
69. 69. A group of 5 boys and 4 girls take part in a study on the type of plants found in a reserved forest area. Each day, two pupils are chosen at random to write a report. • Calculate the probability that both pupils chosen to write the report on the first day are boys. (b) Two boys has written the report on the first day. They are then exempted from writing the report on the second day. Calculate the probability that both pupils chosen to write the report on the second day are the same gender.
70. 70. To choose a boy, the probability, 5 4 P (1 boy ) = st P (2 boy ) = nd 9 8 5 4 5 P (both boys ) = × = 9 8 18
71. 71. Two boys then exempted from writing the report on the second day Both boys-2 boys Both girls-2 girls n(all girls ) n(all boys ) P (one girl ) = P (one boy ) = n(all pupils ) n(all pupils ) 4 3 P (1 girl ) = st P (1 boy ) = st 7 7 3 2 P (2nd girl ) = P (2nd boy ) = 6 6 4 3 2 3 2 1 Thus, P(2boys ) = × = Thus, P(2boys ) = × = 7 6 7 7 6 7 1 2 3 P(both pupils) = + = 7 7 7
72. 72. Society Number of Student boy girl Science 3 5 4 Consumer 6 7 a. If two students were chosen at random from the science society, calculate the probability that both are girls GxG 5 x 4 = 5 8 7 14 b. If two students were chosen at random from the group of boys, calculate the probability that both boys came from the same society S x S or C x C 3/9 x2/8 + 6/9 x 5/8 =1/2
73. 73. 5 marks
74. 74. GRADIENT AND AREA UNDER A GRAPH Distance Speed Constant/ Uniform speed Object stops Time Time m = rate of change of speed m = Rate of change of distance = speed/time = distance/ time = acceleration / deceleration = speed Area under the graph is the distance
75. 75. GRADIENT AND AREA UNDER A GRAPH REMEMBER : 2. Le ng th o f time is to tal time take n 2. AREA o f trape zium 3. Dis tanc e -time g raph , g radie nt = s pe e d e quivale nt to the rate o f c hang e o f dis tanc e 4. S pe e d -time g raph , g radie nt = ac c e le ratio n e quivale nt to the rate o f c hang e o f s pe e d
76. 76. Speed (ms-1) 21 (t ,21) 9 1 (t ,0) O 5 12 t Time (s) (a) State the length of time, in s, that the particle moves with uniform speed. (b) Calculate the rate of change of speed, in ms-1 , in the first 5 seconds.
77. 77. State the length of time, in s, that the particle moves with uniform speed. 12 − 5 = 7 Speed (ms-1) 21 (t ,21) Time (s) 9 1 (t ,0) O 5 12 t
78. 78. -1 (b) Calculate the rate of change of speed, in ms , in the first 5 seconds. Speed (ms-1) 21 (t ,21) 9 1 (t ,0) O 5 12 t Time (s) 9 −1 8 gradient of the green straight line = = 12 − 5 7
79. 79. Calculate the value of t, if the total distance travelled for t seconds is 148 metres. Speed (ms-1) 21 (t ,21) 9 1 (t ,0) O 5 12 t Total distance= area under the graph Stress on the correct values when substituting 1 1 148 = (1 + 9 ) 5 + ( 7 × 9 ) + ( 9 + 21) ( t −12 ) 2 2 10t = 180 t = 18
80. 80. SUGGESTED TIME 15 – 20 MINUTES PER QUESTION
81. 81. 12 (a) Complete table 1 in the answer space for the equation y =2x2-x-3. ( 2 marks) x -2 -1 -0.5 1 2 3 4 4.5 5 y 7 -2 -2 3 12 33 42 12 (a) In the table 1 , find the value of m and the value of n for the equation y = 2 x 3 − 12 x + 3 ( 2 marks) x -3 -2 -1 0 1 2 y -15 m 13 3 -7 n
82. 82. 12 (a) Complete table 1 in the answer space for the equation y =2x2 – x - 3. ( 2 marks) x -2 -1 -0.5 1 2 3 4 4.5 5 y 7 -2 -2 3 12 33 42 0 25
83. 83. 12 GRAPHS OF FUNCTIONS 1. Fill in the blanks in the table y = 2 x − 12 x + 3 3 -2 2 x m n y 11 −5
84. 84. 2. Draw the graphs of functions Scales and the range of x are given −3 ≤ x ≤ 5 Plot the points accurately Can use flexible curve
85. 85. through the curve must be smooth, passing through each point
86. 86. losing marks Using your own scale 12 marks – 1 mark
87. 87. -10 -15 -20
88. 88. e12c. e a n s w e r in t h e a n s w e r s p a c e s p r th x =1.5
89. 89. 12c. The first equation from (a) y = x − 6x + 5 2 The second equation from (c) x = 7x −4 2 eliminate variables that have indices , 2 1 3 x ,x , x
90. 90. 12c. The first equation from (a) y = x − 6x + 5 2 The second equation from (c) x = 7x −4 2 eliminate variables that have indices y − x = x − 6 x + 5 − (7 x − 4) 2 2 y = x − x − 6x − 7x + 5 + 4 2 2 y = −13 x + 9
91. 91. 12d.
92. 92. TRANSFORMATIONS III A Combined Transformation RS means transformation S followed by transformation R. - Use the right terminologies - Start the answer with the right transformation - No short form - Describe in full the transformation – with the correct properties.
93. 93. y E H 4 F y=3 D G 2 Rotation C 180o A B o centre ( 0,3 ) -4 -2 2 4 Describe in full the transformation PQ
94. 94. Enlargement centre( 2,6 ) 6 E H Scale factor 3 4 J M 2 G K 2 4 6 8 -2 L
95. 95. losing marks Please use the right term
96. 96. rotation enlargement correct direction correct centre k : sf : ratio ( m: n) Reflection at a point A enlargement correct scale factor reflection enlargement rotation correct centre enlargement rotation angle translation
97. 97. STATISTICS spm 2006 FREQUENCY POLYGON 2005 HISTOGRAM 2004 HISTOGRAM 2003 OGIVE
98. 98. STATISTICS Mean , median , modal class 2.Frequency Polygon 3.Histogram 4.Ogive 5.Information of the graph
99. 99. Marks Midpoint Frequency 20 – 24 22 5 25 - 29 27 7 30 - 34 32 8 35 - 39 37 10 40 - 44 42 6 45 - 49 47 4 50 - 54 52 2
100. 100. MEAN TABLE x fx 92 92 x 4 = 368 10257 97 97 x 10 = 970 mean = 96 102 102 x 26 = 2652 107 107 x 24 = 2568 112 112 x 17 = 1904 = 106.84 117 117 x 9 = 1053 122 122 x 4 = 488 127 127 x 2 = 254 10257
101. 101. These are students’ answers Mean = 0 x 38 + 4 x 43 + 6 x 48 + 12 x 53 + 9 x 58 + 5 x 63 + 6 x 68 + 8 x 73 0 + 4 + 6 + 12 + 9 + 5 + 6 + 8 = 58.5
102. 102. Frequency polygon table frequency midpoint height f x 90 - 94 4 92 95 - 99 10 97 100 - 104 26 102 105 - 109 24 107 110 - 114 17 112 115 - 119 9 117 120 - 124 4 122 125 - 129 2 127 JUMLAH 96
103. 103. FREQUENCY POLYGON POLIGON KEKERAPAN F 30 R F u ll m a r k s 25 KEKERAPAN E 20 Q U 15 E 10 N C 5 Y 0 87 92 97 102 107 112 117 122 127 132 MIDPOINT TITIK TENGAH
104. 104. POLIGON KEKERAPAN FREQUENCY POLYGON F 30 R E 25 le s s m a r k s KEKERAPAN Q 20 U 15 E N 10 C 5 Y 0 7 7 2 2 2 2 7 87 92 97 12 13 10 10 11 11 12 MIDPOINT MARKAH
105. 105. TABLE FOR HISTOGRAM frequency midpoint Upper boundary CLASS f x 36 – 40 0 38 40.5 41 – 45 4 43 45.5 46 – 50 6 48 50.5 51 – 55 12 53 55.5 56 – 60 10 58 60.5 61 – 65 5 63 65.5 66 – 70 5 68 70.5 71 – 75 8 73 75.5 JUMLAH 50
106. 106. F u ll m a r k
107. 107. 40.5 45.5 50.5 55.5 60.5 65.5 70.5 75.5 F u ll m a r k s
108. 108. 41 - 45 46 - 50 51 - 55 56 - 60 61 - 65 61 - 70 71 - 75 F u ll m a r k
109. 109. Le s s m a r o u ld h a v e g a p
110. 110. 41 - 45 46 - 50 51 - 55 61 - 65 71 - 75 56 - 60 61 - 70 no mark
111. 111. Table for OGIVE frequency Cumulative Upper boundary CLASS f frequency 36 – 40 0 0 40.5 41 – 45 4 4 45.5 46 – 50 6 10 50.5 51 – 55 12 22 55.5 56 – 60 10 32 60.5 61 – 65 5 37 65.5 66 – 70 5 42 70.5 71 – 75 8 50 75.5 ∑ 50
112. 112. Cumulative F u ll m a r k 120 frequency OGIF 100 LONGGOKAN KEKERAPAN 80 60 40 20 0 5 5 5 5 5 5 .5 .5 .5 .5 4. 9. 4. 9. 4. 9. 84 89 94 99 10 10 11 11 12 12 SEMPADAN ATAS Upper boundary
113. 113. Cumulative frequency 120 OGIF 100 LONGGOKAN KEKERAPAN 80 60 le s s m a r k 40 20 0 5 5 5 5 5 5 .5 .5 .5 .5 4. 9. 4. 9. 4. 9. 84 89 94 99 10 11 11 12 10 12 SEMPADAN ATAS Upper boundary
114. 114. Cumulative OGIF frequency N o m a rk 110 100 LONGGOKAN KEKERAPAN 90 80 70 60 50 40 30 20 10 0 0 0 0 0 5 5 5 80 90 85 95 10 10 11 11 12 12 13 limit MARKAH
115. 115. Third Quartile 3 4 1 Median( Second Quartile) 2 1 First Quartile 4
116. 116. Ogive Of Time Taken For 100 Students To Complete Their Compositions 105 100 3 95 100× 90 4 85 80 Finding the third quartile Cumulative Frequency 75 70 65 100 60 55 Finding the median 2 50 (second quartile) 45 40 35 100 30 Finding the first quartile 25 4 20 15 10 5 0 0 20 40 60 80 100 Time ( minutes )
117. 117. INTERQUARTILE Ogive Of Time Taken For 100 Students To Complete RANGE Their Compositions 105 100 Q3 − Q1 3 Q3 95 100× 90 4 85 80 Finding the third quartile Cumulative Frequency 75 70 65 100 60 Finding the median 2 55 50 (second quartile) 45 40 35 Q1 100 30 Finding the first quartile 4 25 20 15 10 5 0 0 20 40 60 80 100 Time ( minutes )
118. 118. Information of the graph 50 students took 60 minutes to complete their composition. Interquartile range is 20 minutes Median is 60 minutes n f o r m a t io n m u s t c o m e f r o m t h e g
119. 119. PLANS AND ELEVATIONS CORRECT SHAPE Satisfy the given CONDITIONS MEASUREMENT MUST BE ACCURATE LATERAL INVERSION is not accepted (SONGSANG SISI TIDAK DITERIMA)
120. 120. 15b(ii). H id d e n lin e
121. 121. losing marks
122. 122. 15 Case 1: Double line Bold line
123. 123. 15a. Case 2 : Sizes – Bigger or Smaller
124. 124. Case 3: extension Case 4 : gap
125. 125. 15 Case 5: Not a right angles
126. 126. Draw a full scale i. The plan of the solid ii. the elevation of the solid as viewed from Y iii. the elevation of the solid as viewed from X X Y
127. 127. F u ll m a r k
128. 128. N o m a rk
129. 129. EARTH AS A SPHERE
130. 130. Longitude N MG 0° 40° 20° 40° E S 60° E
131. 131. TWO meridians form a GREAT circle N 30 w 150 E S
132. 132. U 135 W 45 E S
133. 133. P(60 N, 30 W ) and Q are two points on the surface of the earth where PQ is the diameter of the parallel latitude of P and Q. The position of point Q is A. ( 60 N, 150 W) C. (60 S, 150 E ) B. ( 60 N 150 E) D. (60 S, 150 W ) N P Q 150 E 30 W 60 60 S
134. 134. J( 30 S, 80 E ) and K are two points on the earth where JK is the diameter of the earth. The location of K is A. ( 30 S, 100E) C. ( 30 N, 100W ) B. ( 30 S, 80 E) D. ( 30 N, 80 W) N 80° E 100° W K 300 300 J S
135. 135. a)P is a point on the surface of the earth such that JP is the diameter of the earth. State the position of P. b) Calculate the value of x, if the distance from J to K measured along the meridian is 4200 nautical miles. c)Calculate the value of y, if the distance from J due west to L measured along the common parallel of latitude and then due south to M. f) If the average speed for the whole flight is 600 knots, calculate the time taken for the whole flight.
136. 136. 16. Interpretation of question – sketch the earth o 40 W o J K 140 E o 50 O
137. 137. 16b. JM = MK ∴ angle at the centre JOM = 90 o Greenwich o 40 W o J o K 140 E 40 o 50 M o o 50 E , 50 N
138. 138. 16c. JK = 80 × 60 = 4800 J K o 80 o o 50 50 O
139. 139. 16b. JMK route ∴ angle at the centre JOM = 180 o = 180 × 60 × cos 50 o o 40 B o J 180 K Greenwich M 6942.106 ∴ average speed = 13 = 534
140. 140. Reminders  Calculate , find , solve , – all steps are clearly shown  State – only the answer is required  Unit / label – must be correct if written
141. 141. Reminders Basic Mathematical Skills such as addition, division, subtraction, multiplication. Algebraic and Trigonometric skills Formulae and its applications, Formulae and its substitutions. Round off only at the last answer line.
142. 142. Reminders  Allsteps must be clearly shown.  Read the instructions and questions very carefully .  The answer must be in the lowest form, to 4 significant figures and to 2 decimal places.  Master the calculator  Do not sleep during the exam!
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